$A=(a+b+c+3)\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)$

Bài toán:
Cho$ a;b;c$ thoả mãn $0\leq a\leq b\leq c\leq 1$. Tìm giá trị lớn nhất của biểu thức:$ A=(a+b+c+3)\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)$

Lời giải: Đặt$ x=a+1;y=b+1;c=z+1$

Đưa bài toán về:

$\boxed{Note}$

Cho $x;y;z$ thỏa mãn $1\leq x\leq y\leq z\leq 2$. Tìm Max $P=(x+y+z)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)      (=10)$

Vì $1\leq x\leq y\leq z\leq 2$ nên

$\left(\dfrac{x}{y}-1\right)\left(\dfrac{y}{z}-1\right)\geq 0$

$\Leftrightarrow \dfrac{x}{y}+\dfrac{y}{z}\leq \dfrac{x}{z}+1 \left(\dfrac{y}{x}-1\right)\left(\dfrac{z}{y}-1\right)\geq 0$

$\Leftrightarrow \dfrac{y}{x}+\dfrac{z}{y}\leq \dfrac{z}{x}+1$

Vậy $P\leq 5+2\left(\dfrac{x}{z}+\dfrac{z}{x}\right)$

Mà $\left(\dfrac{x}{z}-1\right)\left(\dfrac{z}{x}-2\right)\geq 0$

$\Leftrightarrow 3\geq \dfrac{2x}{z}+\dfrac{z}{x}$
$\Leftrightarrow \dfrac{x}{z}+\dfrac{z}{x}\leq 3-\dfrac{x}{z}\leq 3-\dfrac{1}{2}=\dfrac{5}{2}$

$\Rightarrow P\leq 10$